Chapter 12 Properties of Solutions. Liquids 2 Properties of Liquids A. Viscosity B. Surface Tension

Chapter 12

Properties of Solutions

Liquids

• 2 Properties of Liquids

• A. Viscosity

• B. Surface Tension

Viscosity

• Viscosity is the resistance of a liquid to flow.

• The stronger the intermolecular forces, the higher the viscosity.

Some Properties of Liquids

Viscosity

Some Properties of Liquids

Surface Tension

Surface Tension• Surface tension is the amount of energy required to

increase the surface area of a liquid. • Resistance to increased surface area or the tendency

of a liquid to maintain minimum surface area.

• Cohesive forces bind molecules to each other.• Adhesive forces bind molecules to a surface.• Ex. miniscus

Some Properties of Liquids

Surface Tension• Meniscus is the shape of the liquid surface.

– If adhesive forces are greater than cohesive forces, the liquid surface is attracted to its container more than the bulk molecules. Therefore, the meniscus is U-shaped (e.g. water in glass).

– If cohesive forces are greater than adhesive forces, the meniscus is curved downwards.

Some Properties of Liquids

Definitions

• Solvent – what does the dissolving–The one that is greater quantity

• Solute – what gets dissolved– can be liquid, solid or gas–The one that is lesser in quantity

A solution is always homogeneous!

Factors Affecting Solubility

• Temperature• Pressure• Solvent-Solvent Interactions

Rule of Thumb

• LIKE DISSOLVES LIKE!

• Polar solutes dissolve in polar solvents.• Non-polar solutes dissolve in non-polar

solvents.

• Compounds with metals (Na, K, Li, etc…) are ionic and therefore polar.

Solution

• Concentration–Molarity (M)–% Mass, % Volume, %Mass/Vol.–Mole Fraction–Molality (m)–Normality (will be skipped)

Molarity (M)

• Molarity = moles of soluteLiter of solution

• Moles = Molarity x Liter of soln.

• Moles = gramsmolar mass

Sample Problem

• Calculate the molarity of a solution containing 3.65 grams of HCl in enough water to make 500.00 mL of solution.

• Ans: 0.200 M

Molality (m)

• Molality = moles of solute kg. solvent

Mole Fraction (c)

• To Solve for the Mole Fraction of A :

cA = Mole AMole A

+ Mole B

Sample Problem

• 1.00 gram of CH3CH2OH (ethanol) is mixed with 100.0 grams of water. Calculate the mole fraction of ethanol in this solution.

3 Types of % Concentrations

–Percent by Mass

–Percent by Volume

–Percent by Mass-Volume

Percent by Mass

• % Mass = mass of solute x 100total mass of solution

• Sample Problem:Find the concentration of 20 grams of

sugar in enough water to make 350 grams of solution.

• Answer: 5.7%

Things to Note:

• Identity of the substance (molecular formula) is not taken into account since only the masses are needed in the equation.

Percent by Volume

• % Volume = volume of solute x 100 Total Vol. of solution

Sample Problem

• 10.0 mL of benzene is added to 40.0 mL of carbon tetrachloride. Find the concentration of the solution.

• Ans: 10 mL / 50 mL) x 100 = 20%

Percent by Mass-Volume

• % Mass-Volume = mass of solute x 100 Total Vol. of Solution

Answers to Problem Set• 1. 357 oC• 2. a. molar mass = 181• b. Tantalum• c. 1.4315 x 10-8

• 3. • 4. a. 0.0247• b. 4.90• c. 0.547

Molality (m)

• Molality = moleskg. solvent

Usefulness of Molality

• Chemists find that molality (m) is a useful concept when they must deal with the effect of a solute on boiling point and freezing point of a solution.

Problem

• A solution is made containing 25.5 g phenol (C6H5OH) in 495 g ethanol (CH3CH2OH) Calculate:

• A. mole fraction of phenol• B. mass % of phenol• C. molality of phenol

Colligative Properties

• Depend upon the number of particles in solution rather than the identity of the compound

• Therefore if solute does not ionize, the concentration of the solution does not change and there is no change in the calculation of the mole fraction (csoln)

Colligative Properties

If solute ionizes, then the number of particles the solute breaks down into should be factored in.

• Example: NaCl Na+ + Cl-

» 1 mole 1 mole 1 mole

Colligative Properties

• If solute is Na2SO4

• Then number of particles is 3• Na2SO4 = 2Na+ + SO4

2-

Colligative Properties

• Boiling Point Elevation• Freezing Point Depression• Osmotic Pressure• Vapor Pressure lowering

Freezing Point Depression

• Presence of solute in solution decreases the freezing point of solutionDT = kfmsolute where

DT = change in Temperature in oC

kf = molal freezing point depression constant (oC/molal)

m = molality

Effect of Ionizing substances

• The presence of solutes that ionize have to be factored into the equation such that if :

DT = kfmsolute , Now: DT = ikfmsolute where i

= the # of ions produced.

Boiling Point Elevation

• Non-volatile solute elevates the boiling point of solvent

• Presence of non-volatile solute decreases vapor pressure, thus solution has to be heated to a higher temperature to boil.

Boiling Point Elevation

DT = kbmsolute

Where:

DT = change in Temp. (in oC)

kb = molal boiling point depression constant (oC/molal)

m = molality

Sample Problem

• A solution is prepared by dissolving 4.9 grams sucrose (C12H22O11 = Mol. Mass: 342.295 g/mol) in 175 grams of water. Calculate the boiling point of this solution. Calculate the freezing point of this solution. Sucrose is a non-electrolyte. Kb of water is 0.51 oC/molal. Kf of water is 1.86 oC/molal.

Sample Problem

• Calculate the boiling point and freezing point of 0.50 m FeCl3 solution. Assume complete dissociation. Kb of water is 0.51 oC/molal. Kf of water is 1.86 oC/molal.

Sample Problem

• A solution of an unknown nonvolatile compound was prepared by dissolving 0.250 g of the substance in 40.0 g of CCl4. The boiling point of the resulting solution was 0.357 oC higher than that of the pure solvent. Calculate the molar mass of the solute. Kb for CCl4 is 5.02. Normal boiling point of CCl4 is 76.8 oC

Lab Experiment

• Determine molality (m) using the equation DT = kfmsolute

• Determine molecular mass of sulfur using the equation

molality (m) = gram mol. mass Kg. solvent

Osmosis

• Osmosis – direction of flow is from less concentrated to more concentrated

- shrivelling

• Reverse Osmosis - from more concentrated to less concentrated

- bloating

Equation

Osmotic Pressure p = MRT whereM = Molarity

R = ideal gas constant (.08206 L-atm/K-mol)

T = Temperature in Kelvin p = osmotic pressure in atm

Effect of Ionizing substances

• Again the presence of solutes that ionize has to be considered:

p = MRT ,

Now: p = iMRT where i = the # of ions

produced.

Sample Problem

• A 0.020 grams sample of a non-dissociating protein is dissolved in water to make 25.0 mL of solution. The osmotic pressure of the solution is 0.56 torr at 25 oC. What is the molar mass of the protein?

Solution Formation

• Energies are involved in solution formation to:

• A. Break the solute apart (endothermic)• B. Break the intermolecular forces of the

solvent (endothermic = DH is positive)• C. Allow solute and solvent molecules to

interact (exothermic = DH is negative)

• Therefore the enthalpy of the solution

DHsolution = DHA + DHB + DHC

• If DHsoln is (+) = means DHC is small

• If DHsoln is (-) = means DHC is large

Factors Affecting Solubility

• A. Structure Effects– Hydrophilic – loves water– Hydrophobic – hates water

• B. Pressure Effects – more important for gases

• C. Temperature effects

IDEAL SOLUTION

• When Raoult’s Law is closely followed• Very little interaction between molecules

(i.e., bonding between solute and solvent)• If only LDF were present, Raoult’s Law

closely followed.

Temperature Effect

• Increasing temperature ONLY increases the RATE of solubility and not solubility itself. In some cases, higher Temp. decreases solubility. Experimentation needed.

Raoult’s Law

• States that the partial pressure exerted by solvent vapor above the solution (PA), equals the product of the mole fraction of the solvent in the solution (CA) times the vapor pressure of the pure solvent (Po)

• PA = (CA)( Po)• Partial pressure = vapor pressure above soln.

Vapor Pressure• The total vapor pressure of the solution is the

sum of the partial pressures of each volatile component.

• Ptot (above soln) = Psolvent(above soln) + Psolute(above soln)

• Partial Pressure of A(solv.) in soln: Psolvent = CsolventPo

• Partial Pressure of B(solute) in soln: Psolute = CsolutePo

Case 1- all volatile components(Ideal Solution)

• The total vapor pressure over the solution is the sum of the partial pressures of each volatile component

• Ptot(above soln) = Psolv. (above soln) + Psolute (above soln)

= (Csolvent in soln)( Po) + (Csolute in

soln)( Po)

Case 2 – non-volatile solute(Non-Ideal Solution)

• Ptot above soln = Psolvent (above soln) + Psolute (above soln)

• Since pure pressure of a non-volatile solute is ZERO, the equation simplifies to:

• Ptot above soln = (CA in soln)( Po)

Sample

• Consider a sample containing 3.0 mole of benzene (C6H6) and 5 moles of toluene (C7H8). The vapor pressure of pure benzene is 88 torr. For pure toluene, it is 34 torr.

• A. What is the mole fraction of benzene? • B. What is the mole fraction of toluene?• C. What are the partial pressures of benzene

and toluene above the solution.• D. What is the total vapor pressure of the

solution?

Problem

• A. Calculate the vapor pressure of water prepared by dissolving 35.0 grams of glycerin (C3H8O3) in 125 grams of water at 343 K. The vapor pressure of pure water is in Appendix B.)

• B. Calculate the mass of ethylene glycol (C2H6O2) that muct be added to 1.00 kg of ethanol (C2H5OH) to reduce its vapor pressure by 10.0 torr at 35 oC. The vapor pressure of pure ethanol at 35 oC is 1.00 x 102 torr.

Problem

• At 20 oC, the vapor pressure of pure benzene is 75 torr and that of pure toluene (C7H8) is 22 torr. Assume that benzene and toluene form an ideal solution.

• A. What is the composition in mole fractions of a solution that has a vapor pressure of 35 torr at 20 oC?

• B. What is the mole fraction of benzene in the vapor above the solution described in part (a)?

Mole fraction in vapor

• C in vapor = P volatile component

P total

HOMEWORK

• Do Problems: 13.59 and 13.61

Henry’s Law

• States that the amount of gas dissolved is proportional to the amount of gas above the solution

Pressure Effect

• Henry’s Law: S = Pk

– Where k is the constant characteristic of a particular solution

– S is the solubility of dissolved gas

• END of CHAPTER 12